Used as the generative function for Bayesian Inferences.
The Binomial Test
Can be used to test if a set of data are equivallent to an expectation given by the underlying probability of seeing an item in “Group 1” (catfish in this example):
\(H_O: p = \hat{p}\)
For this, the test statistics is defined as the number of observations in “Group 1” (e.g., \(N_{catfish}\) in this example), and what we know about the probability of selecting items that have two distinct states is used to evaluate the likelihood of getting \(N_{catfish}\).
The Data
Borrowing from a previous example, let’s assume we are sampling fish in the James River. The hypothesis is that the frequency of catfish is
Exact binomial test
data: samples
number of successes = 52, number of trials = 99, p-value = 5.007e-06
alternative hypothesis: true probability of success is not equal to 0.74
95 percent confidence interval:
0.4223973 0.6265570
sample estimates:
probability of success
0.5252525
Conclusion?
Contained Data
As normal, the object that is returned by the function has all the necessary items for you to include the data into your manuscript.
When we have more than two unique states, we can extend the Binomial test to a more generalized Multinomial expansion. This would be used to test the hypothesis about specific probabilities for \(K-1\) states (the number of independently assorting groups). However, this is a specific version of a more generalilzed approach using Contingency Tables and I won’t be going over it here.
However, there are a few libraries in R available if you’d prefer to use the multinomial expansion.
Contingency Table Tests
Contingency Tables
Contingency tables are another counting approach, where your data can be classified into one or more categories. Often it is taken as a comparison between populations. So, in keeping with our fish theme, let’s assume we went to another river and did another sampling run on catfish.
Nomenclature
This is defined as an \(RxC\) Contingency Table, where \(R\) is the number of Rows and \(C\) is the number of columns.
Underlying Hypothesis
For this, we are assuming that the observations in the row are taken from a larger background population.
Catfish Other
Population 1 52 47
Population 2 32 18
Defining Expectations - No Information
If we have no data, and the underlying hypothesis is that samples may be put into any of the Groups randomly (e.g., the underlying assumption that all fish species have equal abundance), then the expecation is that the frequencies of observations would be \(\frac{1}{K}\), where \(K\) is the number of groups.
Catfish Other
Population 1 73.26 25.74
Population 2 37.00 13.00
Observations and Expectations
The test statistic should measure how close the data in these two matrices are.
Observed
Catfish Other
Population 1 52 47
Population 2 32 18
Expected
Catfish Other
Population 1 73.26 25.74
Population 2 37.00 13.00
Small deviations \(\to\) no differences and failure to reject \(H_O\).
Big deviations \(\to\) rejection of \(H_O\).
Test Statistic
This is based upon an approximation to the \(\chi^2\) distribution, which is well behaved and described for counting data.
\[
T = \sum_{i=1}^r\sum_{j=1}^c\frac{(O_{ij} - E_{ij})^2}{E_{ij}}
\]
Estimating Directly
I’m going to show you how to estimate this from the raw count data so we can get an example where you have to look up the probability for an underlying distribution. There is a chisq.test() function, but you have to be very careful that you have it set up properly for contingency table analyses.
T <-sum( (Observed - Expected)^2/ Expected )T
[1] 26.32813
Rejecting the Null Hypothesis
The general contingency table approach is based upon \(df = (r-1)(c-1)\) independent elements and is distributed as a \(\chi^2\) random variable. In a classic ‘hypothesis testing’ framework with \(\alpha = 0.05\), we would estimate the critical value of the test statistic as:
To get the actual probabilty associated with $T =$26.33, we pass the test statistic to the quantile function for the \(\chi^2\) distribution.
p.value <-1.0-pchisq( T, df=1 )p.value
[1] 2.88063e-07
Tabulating Data
Sample Species Site Measurement
1 1 Catfish Population 1 8.205309
2 2 Catfish Population 1 11.497717
3 3 Catfish Population 1 13.045837
4 4 Catfish Population 1 11.666778
5 5 Catfish Population 1 12.075094
6 6 Catfish Population 1 11.045994
Tabulating Data
The table() function can take the columns of factors and provide for you the observed matrix of counts. If you name your factors descriptively, it will take care of the row and column headers for you.
table( df$Site, df$Species ) -> ObsObs
Catfish Other
Population 1 52 47
Population 2 32 18
Non-Parametric Equivallence
What do we do when our data are not normal or heteroscedastistic?
Non-Parametric Analyses
There are equivalent tests (ok, not equivallent but can be used to answer a similar question) that do not have assumptions of normality, homoscedasticity, etc.
Important
These are largely based upon Ranks and Combinatory Theory.
You gain the applicability with a potential loss of information (and hence statistical power).
Non-Parametric Correlation
The normal cor.test() function has an option to swap out the Pearson-Product Moment approach for one based on Ranks.
Example Comparison
I’m going to use the Iris data set to look at a correlation for relatively poorly behaved data.
There are a few different methods for estimating linear regression models via non-parametric approaches. We’ve seen one common approach whenever we used the stat_smooth() function in ggplot(). By default, it uses the loess() local smoother approach.
The function has a signature similar to that we saw in lm() for normal regression.
lowess Signature
Notice the optional value for span, which defines the size of the local window.
Local Span Size
Fitting A loess Model
fit <-loess( Sepal.Width ~ Sepal.Length, data=iris)fit
Call:
loess(formula = Sepal.Width ~ Sepal.Length, data = iris)
Number of Observations: 150
Equivalent Number of Parameters: 4.74
Residual Standard Error: 0.4172
Examine the Span Parameter
Let’s look at the consequence of the span parameter.
loess.spans <-function(x, y, s){ nobs <-length(y) xs <-sort(x, index.return =TRUE) x <- xs$x y <- y[xs$ix] lo <-loess(y ~ x, span = s)data.frame(Sepal.Length = x, Sepal.Width = lo$fitted, span = s)}
Plotting Different Spans
Optimizing For Span
One method suggested to figure out what an optimal value of span would be is based upon a generalized cross validation (GCV) approach.
There are several additional approaches including Kernel Regression (e.g., Nadaraya & Watson’s Kernel Smoother) and Local Averaging (e.g., Friedman’s Super Smoother). If you need to perform a regression approach, please explore the benefits and challenges with these alternative methods before deciding on which one to use.
Non-Parametric T-Test: Wilcoxon Test
This is an extension of the t-test for the equivalence of the mean values of two data sets (e.g., \(H_O: \bar{x} = \bar{y}\)).
Wilcoxon Test Assumptions
This approach has the following assumptions:
Both sets of data are random samples from their respective population.
Each of the populations are similarly independent.
The measurement scale for each variable is at least ordinal (e.g., you can put them in order).
To test this, we use the Wilcoxon Rank Sum test. The signature of this test is:
Wilcoxon Test
This will take the full set of data, assign them ranked values, and then determine if the ranked values are randomly distributed between the two groups.
fit <-wilcox.test( MPG ~ Transmission, data=data)fit
Wilcoxon rank sum test with continuity correction
data: MPG by Transmission
W = 42, p-value = 0.001871
alternative hypothesis: true location shift is not equal to 0
Contained Values
As for other analyses, the object returned has a number of elements that may be helpful in writing about the analysis.
The Kruskal-Wallis test is an extension on the Wilcoxon test in the same way that the aov() is an extension of the t-test(). The approach is again based upon ranks. Here is an example data set we’ve used once before based on air quality measured across five different months.
